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A continuous function sends sufficiently nearby function inputs to nearby function outputs. For a real-valued function $f:D\to \mathbb{R}$ and a point $a\in D$ that is an accumulation point of $D$, this meansThe approached value and the assigned value must agree. Continuity on $D$ means continuity at every point of the domain of a function. At an endpoint of an interval, only approaches inside that domain of a function count. The epsilon-delta definition of continuity below also applies at an isolated point, where continuity holds automatically.^[ref]

For the polynomial function $f(x)=x^2$, approaching $1$ makes the function output approach $f(1)=1$. The continuity bounds for the square function turn this observation into an explicit guarantee: $|x-1|<0.2$ implies $|x^2-1|<0.6$. This is useful whenever a small error in an function input must produce a controlled error in an function output. The required input tolerance can depend on the point and on the chosen output tolerance.

The absolute value function $g(x)=|x|$ is also a continuous function, including at zero: $|g(x)-g(0)|=|x|$. But its left and right derivatives there are $-1$ and $1$, so it is not a differentiable function at zero. Continuity controls values without requiring a tangent line. Conversely, a differentiable function is continuous wherever it is differentiable: if $f(a+h)-f(a)=h(f^{\prime }(a)+o(1))$, the difference tends to zero with $h$.

Several useful consequences reuse this one local idea. Addition, multiplication and function composition preserve continuity; a quotient of functions is continuous where its denominator is nonzero. On an interval, the intermediate value theorem turns continuity into a guarantee that no intermediate function output is skipped. On a compact set, the extreme value theorem guarantees attained maximum and minimum values. Those extra hypotheses matter: continuity alone on an unbounded domain of a function does not provide a global bound or an attained global maximum.

A rational function such as $x\mapsto 1/x$ is continuous on its domain of a function $\mathbb{R}\setminus \{0\}$. Zero is excluded from that domain of a function; the missing point prevents a continuous extension there, but it does not make the existing map discontinuous at one of its actual inputs. Always state the domain of a function before judging continuity.

For $f:D\to \mathbb{R}$ and $a\in D$, continuity at $a$ is equivalent to the following condition: every sequence $(x_n)$ in $D$ with $x_n\to a$ satisfies $f(x_n)\to f(a)$. This is useful when a numerical method or a proof approaches a point through successively refined function inputs: applying a continuous function preserves the intended limit.

For the forward direction, choose $\delta$ from the epsilon-delta definition of continuity. A convergent sequence eventually lies in that input window, so its function outputs eventually lie in the requested output band. Conversely, if continuity fails, there is some ${\epsilon }_0>0$ for which every input window contains an allowed $x$ with $|f(x)-f(a)|\ge {\epsilon }_0$. Choosing such an $x_n$ within $1/n$ of $a$ gives $x_n\to a$ while $f(x_n)$ does not tend to $f(a)$. This converse relies on the countable shrinking windows available in metric spaces; a general topological space needs a different formulation.

A function is discontinuous at a point in its domain of a function when it fails the epsilon-delta definition of continuity there. Two basic failures are an incorrect assigned value and unequal one-sided limits. They demand different repairs.

For a removable discontinuity, considerThe limit at zero is $0$, but $r(0)=1$. Replacing that one value by $0$ gives the continuous square function. The sequence $x_n=1/n$ makes the mismatch visible: $r(x_n)=1/n^2\to 0\ne r(0)$. A removable discontinuity can also describe an expression with a missing value, in which case the question is whether a continuous extension to the missing input exists.

For a jump discontinuity, take $j(x)=0$ for $x<0$ and $j(x)=1$ for $x\ge 0$. The left and right one-sided limits at zero are $0$ and $1$. No assignment to $j(0)$ can equal both, so changing only the point value cannot make the function continuous there. Contrast both examples with the corner of $|x|$: that function is continuous at zero even though its derivative is absent.